Question

For the network shown below, the source frequency $\omega$ at which applied voltage V and current I are in phase is

• A. 5 rad/sec
• B. 2 rad/sec
• C. 3 rad/sec
• D. 4 rad/sec

Answer 1

The correct option is C 3 rad/sec


For V and I in phase imaginary part of Z(s), should be zero



$Z\left(s\right)=\frac{\frac{1}{s}\times 1}{\frac{1}{s}+1}+0.1s=\frac{1+0.1s+0.1s^2}{s+1}$

Multiplying numerator and denominator by (s - 1)

$Z\left(s\right)=\frac{(0.1s^2+0.1s+1)(s-1)}{(s+1)(s-1)}$

$=\frac{0.1s^3+0.1s^2+s-0.1s^2-0.1s-1}{s^2-1}$

Put $s=j\omega$

$Z\left(j\omega \right)=\frac{0.1(j\omega {)}^3+0.9(j\omega )-1}{(j\omega {)}^2-1}=\frac{j(0.9\omega -0.1{\omega }^3)-1}{-{\omega }^2-1}$

Equating imaginary part to zero,

$0.9\omega -0.1{\omega }^3=0$
${\omega }^2=9$
$\omega =\pm 3$ rad/sec

Alternative Solution:
In frequency domain,


$Z_{th}=\frac{\left(1\right)(-j/\omega )}{1-j/\omega }+0.1j\omega$

$\Rightarrow \frac{j}{j-\omega }+0.1j\omega =0$

$\Rightarrow \frac{j(j+\omega )}{(j-\omega )(j+\omega )}+0.1j\omega =0$

$\Rightarrow \frac{1-j\omega }{1+{\omega }^2}+0.1j\omega =0$

For V and I in phase, imaginary term = 0

Thus, $\frac{-\omega }{1+{\omega }^2}+0.1\omega =0$

$\Rightarrow \omega =3$ rad/s